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GMAT Club Olympics 2025 (Day 8): A family of five has an average

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AVMachine

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09 Jul 2025, 19:35

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A family of five has an average (arithmetic mean) age of 20 years and a median age also 20 years. Each member of the family is a whole number of years old, and all have different ages. Also, the oldest member (o) is 16 years older than the youngest member(y).

o = y+16;

y, A, 20, B, y+16; Actual Ages

20-y; 20-A; 0; 20-B; 20-y-16; Short fall from the average, short fall sum should be zero for avg to be 20

Now, checking the given values, first, the 21 can't be the maximum or minimum value for the oldest member since all have different ages in integers.

1. y+16 = 24; y = 8; short fall 12,20-A,0,20-B,-4; Sum of Short fall 48-A-B=0; A+B=48; Now A can't be greater than 20 and B can't be greater than 24. This can't be the age.

2. y+16 = 26; y = 10; short fall 10,20-A,0,20-B,-6; Sum of Short fall 54-A-B=0; A+B=54; Now A can't be greater than 20 and B can't be greater than 26. This can't be the age.

3. y+16 = 28; y = 8; short fall 12,20-A,0,20-B,-8; Sum of Short fall 44-A-B=0; A+B=44; Now A can't be greater than 20 and B can't be greater than 28. A = 19, B =25; Minumum Value

4. y+16 = 30; y = 14; short fall 6,20-A,0,20-B,-10; Sum of Short fall 36-A-B=0; A+B=36; Now A can't be greater than 20 and less than 14, and B can't be greater than 26 and less than 20. A = 15, B = 21; Maximum Value

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09 Jul 2025, 20:04

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You have to try from the bottom for the minimum and from the top for the maximum.

Oldest Max 30:
14 ? 20 ? 30 -> 15 y 21 would make it possible

Oldest Min 21:
5 ? 20 ? 21 -> You can't fill the 4th place

Oldest Min 24:
8 ? 20 ? 24 -> Any value for the 4th is not enough to bring the average back to 20.

Oldest Min 25:
9 ? 20 ? 25 -> Any value for the 4th is not enough to bring the average back to 20.

Oldest Min 26:
10 ? 20 ? 26 -> 19 and 25 would make it work

Bunuel

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A family of five has an average (arithmetic mean) age of 20 years and a median age also 20 years. Each member of the family is a whole number of years old, and all have different ages. Also, the oldest member is 16 years older than the youngest member.

Select for Minimum the minimum possible age of the oldest member of the family, and select for Maximum the maximum possible age of the oldest member. Make only two selections, one in each column.

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09 Jul 2025, 20:28

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A family of five has an average (arithmetic mean) age of 20 years and a median age also 20 years. Each member of the family is a whole number of years old, and all have different ages. Also, the oldest member is 16 years older than the youngest member.

3rd member = 20

1st member+16= 5th member

20+2nd member+4th member+2*1st mem+16=100
2nd + 4th +2*1st=64

Now for maximize 1st
2nd=19 as 1st & 2nd should be less than 20
4th+2*1st=45
for min minimizing 1st mem we have to maximize 4th member
4th= 1st+15 (one less than 5th member)
3*1st=45-15
1st=10
5th member =26

Now to maximize 1st member we have to minimize 2nd & 4th member

2nd member= 1st member+1
and 4th member=21 just beside median

2*1st+1st+1+21=64
3*1st=42
1st member=14
5th member=14+16=30

Select for Minimum the minimum possible age of the oldest member of the family, and select for Maximum the maximum possible age of the oldest member.

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09 Jul 2025, 20:37

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let's say 5 numbers = a, b, c, d, e; where:

c = 20
e = a + 16
a + b + c + d + e = 100

1. To minimize e, we need to max a , b, d. Let b = c - 1 = 19, e = a +16:
a + 19 + 20 + (a+ 16 - 1)+ (a+ 16) = 100
3a + 70 = 100
a = 10, so e = 26

2. To maximize e, we need to min a , b, d. Let d =21, a = e -16:
e-16 + (e-16-1) + 20 + 21 + e - 100
e =30

answer: Min = 26, max = 30

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09 Jul 2025, 21:21

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"A family of five has an average (arithmetic mean) age of 20 years and a median age also 20 years."

By this information we can get 2 insights:
- If we have an odd number of elements, the median will be the central value. Therefore, the 3rd element (when ordered) has 20 years.
- If the average is 20, the sum of all ages is 5*20 = 100.

"Each member of the family is a whole number of years old, and all have different ages. Also, the oldest member is 16 years older than the youngest member."

Ok. Now we have, ordered, something like:
A, X, 20, Y, A+16 where X + Y + 2A + 16 = 80

"Select for Minimum the minimum possible age of the oldest member of the family, and select for Maximum the maximum possible age of the oldest member. Make only two selections, one in each column."

To maximize a number, we normally need to minimize the other numbers. In this case, however, we have a constraint about the lowest value: it needs to be only 16 lower than the maximum. But we can minimize the "Y" and try to minimize "X" so we can use the values between A and A+16.

Let's try values: the greatest option is 30. If we plug it, we'll have:
(30-16), X, 20, Y, 30
14, X, 20, Y, 30
X + Y = 100 - 64 = 36
If we set X = 14 + 1 = 15, we have 21 to set as Y. This fits perfectly:
14, 15, 20, 21, 30

So, 30 is the maximum possible value.

To minimum, let's try 25 (an intermediate option, so if it doesn't work we cut out half of the possibilities):
(25-16), X, 20, Y, 25
9, X, 20, Y, 25
X + Y = 100 - 54 = 46
If we set X = 20 - 1 = 19 (the maximum X value), we need to have Y = 46 - 19 = 27, that is greater than 25. So, we can not have a valid set with 25 as the oldest member.

Let's try 26 now:
(26-10), X, 20, Y, 26
10, X, 20, Y, 26
X + Y = 100 - 56 = 44
If we set X = 20 - 1 = 19 (the maximum X value), we need to have Y = 44 - 19 = 25, that is smaller than 26. So, we can have a valid set with 26 as the oldest member.

Therefore:

Answers =
Minimum: 26
Maximum: 30

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09 Jul 2025, 21:31

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NOTES:
Family of 5
Average/Arithmetic mean = 20 years
Median age = 20 years
Each member is a whole number of years
All ages different
Oldest is 16 years more than youngest.
All years combined =100

SOLVE:
Ages:
X1, X2, 20, X3, X1+16 (Sum is 100)

Lets plug in options for oldest. then see if X2, and X3 can work to add to difference

21 - does not work since X3 needs a different age between 20-21
24 - 8, X2, 20, X3, 24 = X2+X3 +52. | X2+X3 =48 | 48-19 = 29 | 48-23 = 25 Does not work
25 - 9, X2, 20, X3, 25 = X2+X3+54 | X2+ X3 = 46 | 46-19 =27 | 46-24 =22 does not work
26 - 10, X2, 20, X3, 26 = X2+X3+56 | X2 + X3 = 48 | 48-19 = 29 | 48-25 = 23 does not work
28 - 12, X2, 20, X3, 28 = X2+X3 + 60 | X2+X3 = 40 | 40-19= 21 | 40-27 = 13| WORKS!
30 - 14, X2, 20, X3, 30 = X2 + X3 +64 | X2+ X3 = 36 | 36-19 = 17 | 36-15 = 21| WORKS!

MIN- 28
MAX- 30

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09 Jul 2025, 21:43

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Average age = 20.
So sum of ages = 100
Median = 20
let the youngest age be x
ages of family members = x, _ , 20 , _ , x+16

keep in mind that ages are all different and whole numbers.

Using trial and error:

minimum age = 28 and maximum age = 30

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Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

A family of five has an average (arithmetic mean) age of 20 years and a median age also 20 years. Each member of the family is a whole number of years old, and all have different ages. Also, the oldest member is 16 years older than the youngest member.

Select for Minimum the minimum possible age of the oldest member of the family, and select for Maximum the maximum possible age of the oldest member. Make only two selections, one in each column.

So answer is 28 and 30.

Let's think age of youngest be y and oldest be y+16. and since mean and median are same which meand 3rd member age is 20. so now we have to think about age of 2nd and 4th member so that the oldest have minimum and maximum age.

Therefore, it is possible when all the number in middle which means 2nd, 3rd and 4th are closer to each other so lets say they are 19,20 and 21. this gives us age of older one be 28. Now, to create a minimum value for oldest person we have when the 2nd person age is also close to lowest and 4th person age is close to mean so we can create a assumtion when age are in like this: y, y+1, 20,21, y+16 . so here we get y+16 =30 which is his maximum age.

so right answers are 28 and 30

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09 Jul 2025, 22:55

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The total age sum is 5×20=100.

Median age, a3= 20
Oldest age, a5=a1+16
Substitute a3 and a5 into the sum: 2a1+a2+A4=64
Since ages are distinct and ordered , a2 must be less than 20 and a4 between 21 and a1+15
a2+a4=64-2a1
a5=a1+16
range from 26(where a1=10) upto 30 where a1 is 14 giving minimum and maximum possible oldest ages.

Answer: Minimum oldest age is 26 and maximum oldest age is 30.

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Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

A family of five has an average (arithmetic mean) age of 20 years and a median age also 20 years. Each member of the family is a whole number of years old, and all have different ages. Also, the oldest member is 16 years older than the youngest member.

Select for Minimum the minimum possible age of the oldest member of the family, and select for Maximum the maximum possible age of the oldest member. Make only two selections, one in each column.

A family of five has an average (arithmetic mean) age of 20 years and a median age also 20 years.
X = 20
M = 20

Total = 100

Let the members (in increasing order of age)be
A, B, C, D, E
=> C = 20

Each member of the family is a whole number of years old
All have different ages

Also, the oldest member is 16 years older than the youngest member.
=> D = A + 16

Also
A + B + 20 + D + A + 16 = 100
=> 2A + B + D = 64

1. To calculate the minimum age of E we need to maximize the age of all others
=> 2A + B + D = 64
=> 2A + 19 + A + 15 = 64 ----- [ D is taken only 1 less than E and B is taken only 1 less than C ]
=> 3A = 64 - 34
=> A = 30/3 = 10
Hence E = 16 + 10 = 26

1. To calculate the maximum age of E we need to minimize the age of all others
2A + B + D = 64
=> 2A + A + 1 + 21 = 64 ------- [ D is taken only 1 more than C and B is taken only 1 more than A ]
=> 3A = 64 - 22
=> 3A = 42
=> A = 14
Hence E = 16 + 14 = 30

Minimum = 26
Maximum = 30

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10 Jul 2025, 00:09

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We know the family has 5 members with:

An average age of 20 → total age = 5 × 20 = 100.

A median age of 20 → the middle value when ages are listed in order is 20.

All ages are distinct whole numbers.

The oldest is 16 years older than the youngest.

Let’s call the ages (in order): a < b < c < d < e, with c = 20 and e = a + 16.

To find the minimum oldest age (e):
We want to keep the oldest as young as possible, so we set the youngest (a) as low as we can while ensuring all ages are distinct and add up to 100.

Try a = 10, then e = 10 + 16 = 26.
Possible ages: 10, 19, 20, 25, 26 → All distinct, median is 20, total is 100
So the minimum oldest age is 26.

To find the maximum oldest age (e):
To make the oldest as old as possible, set the youngest (a) higher. Try a = 12, so e = 12 + 16 = 28.
Possible ages: 12, 16, 20, 24, 28 → All distinct, median 20, total 100
So the maximum oldest age is 28.

Final Answer:

Minimum = 26

Maximum = 28

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10 Jul 2025, 00:16

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The ages of five family members $- a, b, 20, c, d$ & $a+b+c+d=80$

$a=d-16$, so $2d+b+c=96$

Minimum:

To minimize $d$, all other values need to be maximum.

Max values of $b=19$ & $c=d-1$

$2d+19+d-1=96$; $3d+18=96$

$d=26$

Maximum:

To maximize $d$, all other values need to be minimum.

Min values of $b=a+1=d-15$ & $c=21$

$2d+d-15+21=96$; $3d+6=96$

$d=30$

Minimum: 26
Maximum: 30

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Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

A family of five has an average (arithmetic mean) age of 20 years and a median age also 20 years. Each member of the family is a whole number of years old, and all have different ages. Also, the oldest member is 16 years older than the youngest member.

Select for Minimum the minimum possible age of the oldest member of the family, and select for Maximum the maximum possible age of the oldest member. Make only two selections, one in each column.

Average of the age of 5 members = 20 years ---> Total age of 5 members = 5*20= 100 years.

Let the age of 5 members be a, b, 20 , c, a+16 years in ascending order.

a<b<20<c<a+16

For minimum possible age of the oldest, ages of other members has to be maximum:
maximum b=19, maximum c = a+15

Total Age = a+b+20+c+a+16 = 100
a+19+20+a+15+a+16=100
3a+70=100
a=10
Minimum possible age of the oldest member of the family = 26 years.

For maximum possible age of the oldest, ages of other members has to be minimum:
minimum b=a+1, minimum c = 21

Total Age = a+b+20+c+a+16 = 100
a+a+1+20+21+a+16=100
3a+58=100
3a=42
a=14
Maximum possible age of the oldest member of the family = 30 years.

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A.M. = 20
Sum = 100
Median = 20
Let the age of oldest member = x
Youngest member = x-16
The ascending order would be : x-16, y , 20, z , x
y & z are the remaining two members.
we know that sum would be 100
x-16 < y < 20
& 20< z < x

So keeping the above conditions and checking the value of x as given in the column .
x (min.) = 26
x (max.) = 30

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Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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A family of five has an average (arithmetic mean) age of 20 years and a median age also 20 years. Each member of the family is a whole number of years old, and all have different ages. Also, the oldest member is 16 years older than the youngest member.

Select for Minimum the minimum possible age of the oldest member of the family, and select for Maximum the maximum possible age of the oldest member. Make only two selections, one in each column.

Let’s denote the five distinct whole-number ages in increasing order:
a<b<20<d<a + 16
We have: a + b + 20 + d + (a+16) = 20*5 = 100
=> 2*a + b + d = 64
1. a min when b and d max => b=19 and d = a + 15 => 2*a + 19 + (a+15) = 64 => a = 10
So, the minimum possible age of the oldest member of the family is 10+16 = 26
2. a max when b and d min => b = a + 1 and d=21 => 2*a + (a+1) + 21 = 64 => 14
So, the maximum possible age of the oldest member of the family is 14+16 = 30

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10 Jul 2025, 03:25

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a<b<c<d<e; c=20

We want to minimise a, so e = a + 16 is also small

Let b=c-1=19
and d=a+15
Thus a+19+20+a+15+a+16=100
3a+70=100
3a=30
a=10
Thus e=26
10,19,20,25,26

We want to maximise a, so e = a + 16 is also large

Try:
a =14, e= 30
2a+b+d = 64
b + d =36

Try b =15, d = 21
14, 15, 20, 21, 30
e= 30

Ans 26,30

Lemniscate

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10 Jul 2025, 04:23

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ages:
a<b<20<c<d=a+16

a+b+20+c+a+16 = 20*5 = 100
2a+b+c = 64

Minimum when b and c are maximum:
2a+19+a+15 = 64
3a = 30
a = 10
minimum = 10+16 = 26

Maximum when b and c are minimum:
2a+a+1+21 = 64
3a = 42
a = 14
maximum = 14+16 = 30

Minimum=26
Maximum=30

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Avg is 20 for 5 people, i.e 5*20=100
median is 20 , so the third member is having age 20.the oldest is 16 years older than younger one.

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A<B<20<C<A+16

To minimize A+16, we need to minimize A maximizing B and C
A + 19 + 20 + A+15 + A+16 = 20*5 = 100
3A = 30
A = 10 -> (10,19,20,25,26) -> minimum=26

To maximize A+16, we need to maximize A minimizing B and C
A + A+1 + 20 + 21 + A+16 = 20*5 = 100
3A = 42
A = 14 -> (14,15,20,21,30) -> maximum=30

Minimum=26 and Maximum=30

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Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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A family of five has an average (arithmetic mean) age of 20 years and a median age also 20 years. Each member of the family is a whole number of years old, and all have different ages. Also, the oldest member is 16 years older than the youngest member.

Select for Minimum the minimum possible age of the oldest member of the family, and select for Maximum the maximum possible age of the oldest member. Make only two selections, one in each column.